Question

How do you find the critical points for `f(x)= (sinx) / (cosx-2)` and the local max and min?

Answer 1

`f(x)` has a maximum in `x=-pi/3+2kpi` and a minimum in `x=pi/3+2xpi`

Explanation:

Critical points are where the first derivative is zero.

`f(x) = frac sinx (cosx-2)`

`f'(x) = frac ( (cosx -2) cosx + sin^2x) ((cosx-2)^2) = frac ( cos^2x -2 cosx + sin^2x) ((cosx-2)^2) = (1-2cosx) / (cosx-2)^2`

So critical points are where:

`1-2cosx=0`

`cosx=1/2`

`x=pi/3+2kpi` and `x=-pi/3+2kpi`

Without calculating the second derivative, we can see that the denominator of `f'(x)` is always positive, so the sign of the derivative is the sign of `1-2cosx`.

So, for `x_k=pi/3+2kpi`, `f'(x_k-epsilon) <0` and `f'(x_k+epsilon) >0` so these points are minimums, while for `x_k=--pi/3+2pi` the opposite is true and these points are maximums.

graph{sinx/(cosx-2) [-10, 10, -5, 5]}